\begin{example}

Consider a causal LTI system given by the recursive formula:

$$
y(n) - \alpha y(n-1) = x(n) \quad \abs{\alpha} \lt 1
$$

Suppose that $x(n) = u(n)$. Determine $y(n)$. You will find that $y(n) = y_{tr}(n) + y_{ss}(n)$, where  $y_{tr}(n)$ is the transient term, and $y_{ss}(n)$ is the steady-state term. 

\begin{align*}
y(n) - \alpha y(n-1) &= x(n) \\
\hat{Y}(z) - \alpha z^{-1} \hat{Y}(z) &= \hat{X}(z) \\
\hat{Y}(z)(1 - \alpha z^{-1}) &= \hat{X}(z) \\
\hat{H}(z) &= \frac{1}{(1 - \alpha z^{-1})} \\
\end{align*}

We know that $\hat{Y}(z) = \hat{H}(z)\hat{X}(z)$ and $\hat{X}(z) = \frac{1}{1 - z^{-1}}$.

\begin{align*}
\hat{Y}(z) &= \hat{H}(z)\hat{X}(z) \\
\hat{Y}(z) &= \lr{\frac{1}{1 - \alpha z^{-1}}} \lr{\frac{1}{1 - z^{-1}}} \\
\hat{Y}(z) &= \frac{z^2}{\lr{ z - 1} \lr{ z - \alpha }} \\
\end{align*}

We know that $RoC(h) = \{ z \st \abs{\alpha} \lt \abs{z} \}$, $RoC(x) = \{ z \st \abs{z} \gt 1 \}$. The $RoC(y)$ is the intersection of these two regions, hence $RoC(y) = \{ z \st \abs{z} \gt 1 \}$. Now we can break $\hat{Y}(z)$ into parts to determine the inverse transform.

\begin{align*}
\hat{Y}(z) &= \frac{z^2}{\lr{ z - 1} \lr{ z - \alpha }} \\
\frac{\hat{Y}(z)}{z} &= \frac{z}{\lr{ z - 1} \lr{ z - \alpha }} \\
\frac{\hat{Y}(z)}{z} &= \frac{A}{z - 1} + \frac{B}{z-\alpha} \\
\end{align*}

Using the Heaviside method we can find $A$ and $B$:

\begin{align*}
A &= \left. \frac{z}{z-\alpha}\right|_{z = 1} = \frac{1}{1-\alpha} \\
B &= \left. \frac{z}{z-1}\right|_{z = \alpha} = \frac{\alpha}{\alpha - 1} \\
\end{align*}

Now we can solve for the inverse Z-transform:
\begin{align*}
\frac{\hat{Y}(z)}{z} &= \frac{\frac{1}{1-\alpha}}{z - 1} + \frac{\frac{\alpha}{1-\alpha}}{z-\alpha} \\
z^{-1}\hat{Y}(z) &= \frac{\frac{1}{1-\alpha}z^{-1}}{1 - z^{-1}} + \frac{\frac{\alpha}{1-\alpha}z^{-1}}{1-\alpha z^{-1}} \\
\hat{Y}(z) &= \frac{\frac{1}{1-\alpha}}{1 - z^{-1}} + \frac{\frac{\alpha}{1-\alpha}}{1-\alpha z^{-1}} \\
y(n) &= \frac{1}{1-\alpha} u(n) + \frac{\alpha}{1-\alpha} \alpha^n u(n) \\
\end{align*}

Note that $y_{tr}(n) = \frac{1}{1-\alpha} u(n)$ since it emanates from the input. As $n \to \infty$, $y_{tr}(n) \to 0$. The steady state function $y_{ss}(n) = \frac{\alpha}{1-\alpha}\alpha^n u(n)$ because it emanates from the system. As $n \to \infty$, $y_{ss}(n) \to \frac{1}{1-\alpha}$. If $x(n) = 1$ instead of $u(n)$, note that the output is equal the limit of the steady state function as $n$ approaches infinity: $\lim_{n\to\infty} y_{ss}(n) = \frac{1}{1-\alpha}$.

\begin{align*}
y(n) &= z_0 \hat{H}(z) = \frac{z_0}{1 - \alpha z^{-1}} \\
y(n) &= 1^n \hat{H}(1) = \frac{1}{1-\alpha} \\
\end{align*}

In general if you have a BIBO stable system and pass an input $x(n)$ through it, you will have a term due to the system pole and a term due to the suddenly applied input. Consider two simplified cases, one where $x(n) = 1$ and another where $x(n) = u(n)$. After $n$ approaches infinity, the responses are indistinguishable. What if $x(n) = \beta^n$?



I) $\abs{\alpha} \lt \abs{\beta}$ 

\begin{align*}
y(n) &= k_1 \beta^n + k_2 \alpha^n u(n) \\
\end{align*}

$\beta^n$ persists longer as $n$ approaches $\infty$. How can you find $k_1$?

\begin{align*}
k_1 &= \left. \hat{H}(z) \right|_{z = \beta} = \frac{\beta}{\beta - \alpha} \\
\end{align*}

We can also determine the coefficient through the Z domain representation:

\begin{align*}
\hat{Y}(z) &= \hat{H}(z) \hat{X}(z) \\
\hat{Y}(z) &= \frac{z}{z - \alpha} \frac{z}{z - \beta} \\
\hat{Y}(z) &= \frac{A}{z - \alpha}  + \frac{B}{z - \beta} \\
\hat{Y}(z) &= \frac{k_2}{z - \alpha}  + \frac{\frac{\beta}{\beta - \alpha}}{z - \beta} \\
\end{align*}

\end{example}


Steady-State